Two lenses of power `+ 10 D and - 5 D` are placed in contact,
(i) Calculate the focal length of the combination
(ii) where should an object be held from the combination so as to obtain a virtual image of magnification `2` ?

Options

A. `5 cm`
B. `-5 cm`
C. `10 cm`
D. `-10 cm`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Power of the Combination of Lenses
The power of a lens is given in diopters (D), and the total power \( P \) of two lenses in contact is the sum of their individual powers.

Given:
- Power of lens 1, \( P_1 = +10 \, D \)
- Power of lens 2, \( P_2 = -5 \, D \)

The formula for the total power \( P \) is:
\[ P = P_1 + P_2 \]

Substituting the values:
\[ P = 10 - 5 = 5 \, D \]

### Step 2: Calculate the Focal Length of the Combination
The focal length \( f \) of a lens is related to its power \( P \) by the formula:
\[ P = \frac{1}{f} \]

Rearranging this gives:
\[ f = \frac{1}{P} \]

Substituting the value of \( P \):
\[ f = \frac{1}{5} = 0.2 \, m \]

Thus, the focal length of the combination is \( 0.2 \, m \) or \( 20 \, cm \).

### Step 3: Find the Object Distance for a Virtual Image with Magnification of 2
The magnification \( m \) is given by the formula:
\[ m = \frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance. Since we want a virtual image, the magnification will be negative. Given:
\[ m = -2 \]

This implies:
\[ -2 = \frac{v}{u} \implies v = -2u \]

### Step 4: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Substituting \( v = -2u \) into the lens formula:
\[ \frac{1}{f} = \frac{1}{-2u} - \frac{1}{u} \]

Finding a common denominator:
\[ \frac{1}{f} = \frac{-1}{2u} - \frac{2}{2u} = \frac{-3}{2u} \]

### Step 5: Substitute the Focal Length
We already calculated \( f = 0.2 \, m \), so substituting this into the equation:
\[ \frac{1}{0.2} = \frac{-3}{2u} \]

Calculating \( \frac{1}{0.2} \):
\[ 5 = \frac{-3}{2u} \]

### Step 6: Solve for \( u \)
Rearranging gives:
\[ 2u \cdot 5 = -3 \implies 10u = -3 \implies u = \frac{-3}{10} = -0.3 \, m \]

Thus, the object should be placed at \( 0.3 \, m \) (or \( 30 \, cm \)) from the lens combination.

### Summary of Results
1. The focal length of the combination is \( 0.2 \, m \) or \( 20 \, cm \).
2. The object should be placed at \( 0.3 \, m \) or \( 30 \, cm \) from the lens combination to obtain a virtual image with a magnification of \( 2 \).